OT_Cercles / Tractors

To find an orthogonal trajectory, we generally replace the slope with its negative inverse in its differential equation of origin (DE). That is to say in replacement $ phi rightarrow pi / 2- phi $

The EDs of the meridians of a sphere and a pseudosphere are respectively:

$$ cos phi = r / a; quad sin phi = r / a; $$

To plot meridians, EDs can be easily expressed as a product of major curvatures (initiated on $ z $-axis variable).

$$ dfrac {r ^ {}} {r (1 + r ^ {2}) ^ 2} = pm 1 $$

I've tried drawing the two meridians with NDSolve by choosing an orthogonal intersection at the starting point.

```
Clear["`.*"]
a = 1; sgn = -1; zmax = 1.5999;
ri = 0.8; slp = Sqrt[1 - ri^2]/ ri; ar = 0.5;
NDSolve[{R''[z]/ R[z]/ (1 + R & # 39;[z]^ 2) ^ 2 == sgn 1 / a ^ 2, R & # 39;[0] == slp,
R[0] == ri}, R, {z, 0, zmax}
;
r[t_] = R
sph = Parcel[{R[{R[{r[{r[z], 0}, {z, 0, zmax}, PlotStyle -> {Red, Thick},
PlotLabel -> "SPH", AspectRatio -> ar, GridLines -> Automatic];
"~~~ sign change of Gauss curl ~~~"
sgn = +1;
NDSolve[{R''[z]/ R[z]/ (1 + R & # 39;[z]^ 2) ^ 2 == sgn 1 / a ^ 2, R[0] == laughed,
R & # 39;[0] == -1 / slp}, R, {z, 0, zmax}
;
r[t_] = R
ps = plot[{R[{R[{r[{r[z], 0}, {z, 0, zmax 1.1}, PlotStyle -> {Blue, Thick},
PlotLabel -> "PS & SPH", AspectRatio -> ar, GridLines -> Automatic];
Show[{ps, sph}, PlotRange -> All]
```

However, the second intersection does not seem to be orthogonal convincingly.

I can not trace my mistake. Thank you in advance for your attention and I will explain further what may not be clear enough.